1868_United_States_presidential_election_in_Rhode_Island Knowpia

1868 United States presidential election in Rhode Island
County Results Grant 60-70% 70-80%
President before election; Andrew Johnson; Democratic	Elected President ; Ulysses S. Grant; Republican

The 1868 United States presidential election in Rhode Island took place on November 3, 1868, as part of the 1868 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Rhode Island voted for the Republican nominee, Ulysses S. Grant, over the Democratic nominee, Horatio Seymour. Grant won the state by a margin of 32.98%.

With 66.49% of the popular vote, Rhode Island would be Grant's fifth strongest victory in terms of popular vote percentage after Vermont, Massachusetts, Kansas and Tennessee.^[1]

Results edit

1868 United States presidential election in Rhode Island^[2]
Party		Candidate	Running mate	Popular vote		Electoral vote
Party		Candidate	Running mate	Count	%	Count	%
	Republican	Ulysses S. Grant of Illinois	Schuyler Colfax of Indiana	12,993	66.49%	4	100.00%
	Democratic	Horatio Seymour of New York	Francis Preston Blair, Jr. of Missouri	6,548	33.51%	0	0.00%
Total				19,541	100.00%	4	100.00%

References edit

^ "1868 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
^ "1868 Presidential General Election Results - Rhode Island".

[1] "1868 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.

[results-2] "1868 Presidential General Election Results - Rhode Island".

[1]

[2]

Summary

Results edit

See also edit

References edit